A function $f$ on $X$ is said to separate points of $X$ if for $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$. Suppose $A$ is a subset of $Y^{*}$, the space of functionals on a Banach space $Y$, which separates points. $T$ is a linear map on $X$ to $Y$. Prove that $T$ is continuous if and only if $g\circ T$ is continuous for all $g\in A$.
Any hints will be appreciated.
Thanks in advance!
Here is a proof when $X$ is a F-space(this is needed to apply the closed graph theorem)
If $T$ is continuous, then $g \circ T$ is continuous as composition of continuous mappings.
If $g\circ T$ is continuous, we are going to apply closed graph theorem to prove $T$ is continuous:
Suppose $x_n \to x$ in $X$ and $Tx_n \to y$ in $Y$, then we have $g\circ T(x-x_n) \to 0$ since $g\circ T$ is continuous, which is to say $g\circ Tx_n \to g\circ Tx$, i.e $$g(Tx_n) \to g(Tx)$$
Since $g$ is continuous and $Tx_n \to y$ in $Y$, we have $$g(Tx_n) \to g(y)$$
So we have $g(y)= g(Tx)$. Since $A$ separates points, we must have $y = Tx$.
So the graph $\{(x, Tx), x \in X\}$ in closed in $X \times Y$ and the closed graph theorem says $T$ is continuous